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Byju's Answer

Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

When 3.00 g o...

Question

When $3.00 g$ of a substance $X$ is dissolved in $100 g$ of $C C l_{4}$ , it raises the boiling point by 0.60 K. The molar mass of the substance $X$ in $g {mol}^{- 1}$ . (Nearest integer) is
[Given $K_{b}$ for $C C l_{4}$ is $5.0 K k g {mol}^{- 1}$ ]

250.00

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250.0

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250

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Solution

$Δ T_{b} = K_{b} m$
$m$ is molality
$Δ T_{b} = 0.6$ is elevation in boiling point
$K_{b} = 5$ is ebullioscopic constant

$0.60 = \frac{5.0 \times 3.00 \times 1000}{M \times 100}$

$M = 250$

Molecular weight of the substance = 250

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Similar questions

The addition of

3 g

of a substance to

100 g C C l_{4} (M = 154 g m o l^{- 1}

) raises the boiling point of

C C l_{4}

{0.60}^{\circ} C

. If

K_{b} (C C l_{4})

5 K m o l^{- 1} k g

, calculate the molar mass of the substance.

Given:

K_{f} (C C l_{4}) = 31.8 K k g m o l^{- 1}

ρ (s o l u t i o n) = 1.64 g c m^{- 3}

.
(Answer in g/mol)

Q. The addition of 3 gm of substance to 100 gm

C C l_{4} (M = 154 g m m o l^{- 1})

raises the boiling point of

C C l_{4}

{0.60}^{\circ} C K_{b}

C C l_{4}

is 5.03 kg

m o l^{- 1}

Calculate :

(a) the freezing point depression

(b) the relative lowering of vapour pressure

(d) the molar mass of the substance

[Given :

K_{f} (C C l_{4}) = 31.8 k g m o l^{- 1} K

and density of solution =

1.64 g m / c m^{3}

]

Q. The boiling point of benzene is

353.23 K

. When

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of a non-volatile solute was dissolved in

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of benzene, the boiling point is raised to

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. Calculate the molar mass of the solute.

K_{b}

for benzene is

2.53 K k g m o l^{- 1}

Q. The boiling point of a solution containing

68.4

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= 342

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) in

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[

K_{b}

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= 0.512

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Q. When

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When 3.00 g of a substance X is dissolved in 100 g of CCl4, it raises the boiling point by 0.60 K. The molar mass of the substance X in g mol−1. (Nearest integer) is [Given Kb for CCl4 is 5.0 K kg mol−1 ] ​​​​​​​

ΔTb=Kb m m is molality ΔTb=0.6 is elevation in boiling point Kb=5 is ebullioscopic constant 0.60=5.0×3.00×1000M×100 M=250 Molecular weight of the substance = 250

When $3.00 g$ of a substance $X$ is dissolved in $100 g$ of $C C l_{4}$ , it raises the boiling point by 0.60 K. The molar mass of the substance $X$ in $g {mol}^{- 1}$ . (Nearest integer) is
[Given $K_{b}$ for $C C l_{4}$ is $5.0 K k g {mol}^{- 1}$ ]

$Δ T_{b} = K_{b} m$
$m$ is molality
$Δ T_{b} = 0.6$ is elevation in boiling point
$K_{b} = 5$ is ebullioscopic constant

$0.60 = \frac{5.0 \times 3.00 \times 1000}{M \times 100}$

$M = 250$

Molecular weight of the substance = 250